Previous Page 20 of 28 Next → The equilibrium constant for the following reaction is 77.3 at a certain temperature. 4 A(g) + B₂(g) 3 C(g) 0.100 mol of A and 0.335 mol of B are present at equilibrium in a 5.00-L flask. What is the molar concentration of C in the flask at equilibrium? 61°F Clear Esc O A. 9.39×10 M OB. 4.26 M OC. 0.470 M O D. 8.29x10-7 M. O E. 0.137 M ⒸN F3 #3 - A F4 $ References Use the References to access important values if needed for this question. 54 % 5 QL F6 DELL FB & 7 7 F10 F11 11 F12

Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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All the back and forwth in these assignments. I am not keeping my formulas stright. Please help
ck
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The equilibrium constant for the following reaction is 77.3 at a certain temperature.
4 A(g) + B₂(g) =3 C(g)
0.100 mol of A and 0.335 mol of B are present at equilibrium in a 5.00-L flask. What is the molar concentration of C in the flask at equilibrium?
61°F
Clear
Esc
OA. 9.39x103³ M
B. 4.26 M.
OC. 0.470 M
O D. 8.29x10-7 M
O E. 0.137 M
Q
A
Z
@
2
W
S
X
Alt
F3
3
E
D
h
F4
4x
$$
References
Use the References to access important values if needed for this question.
4
C
F5
R
F
▬▬▬
8
%
5
V
QL
F6
T
8
G
DELL
B
FB
&
7
H
7
FO
*
4
N
8
d
8
O
F10
SAK
K
M
O
F11
2
0
50
6
F12
Transcribed Image Text:ck Previous Page 20 of 28 Next → The equilibrium constant for the following reaction is 77.3 at a certain temperature. 4 A(g) + B₂(g) =3 C(g) 0.100 mol of A and 0.335 mol of B are present at equilibrium in a 5.00-L flask. What is the molar concentration of C in the flask at equilibrium? 61°F Clear Esc OA. 9.39x103³ M B. 4.26 M. OC. 0.470 M O D. 8.29x10-7 M O E. 0.137 M Q A Z @ 2 W S X Alt F3 3 E D h F4 4x $$ References Use the References to access important values if needed for this question. 4 C F5 R F ▬▬▬ 8 % 5 V QL F6 T 8 G DELL B FB & 7 H 7 FO * 4 N 8 d 8 O F10 SAK K M O F11 2 0 50 6 F12
Expert Solution
Step 1

Given -

4A(g)+B2(g) <->3C(g).  K = 77.3 

Moles of A= 0.100 mol 

Moles of B2 = 0.335 mole ( There is no B hence there is mole of B2

Volume=5.00L 

 

 

 

 

 

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